Question

Prove the following trigonometric identities.

`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = 2 sec theta`

Answer 1

We have to prove `(1 + sin theta)/cos theta + cos theta/1+ sin theta - 2 sec theta`

We know that, `sin^2 theta + cos^2 theta = 1`

Multiplying the denominator and numerator of the second term by `(1 - sin theta)` we have

`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = (1 = sin theta)/cos theta =  (cos theta(1 - sin theta))/((1 + sin theta)(1 - sin theta))`

`= (1 + sin theta)/cos theta =  (cos theta (1 - sin theta))/(1-sin theta)`

`= (1 + sin theta)/cos theta + (cos theta(1 - sin theta))/cos^2 theta`

`= (1 + sin theta)/cos theta + (1 - sin theta)/cos theta`

`= (1 + sin theta +  1 -sin theta)/cos theta`

`= 2/cos theta`

`= 2 sec theta`

Answer 2

LHS = `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`

= `(( 1 + sin θ)^2 + cos^2 θ)/(cos θ( 1 + sin θ))`

= `( 1 + sin^2 θ + 2 sin θ + cos^2 θ)/(cos θ( 1 + sin θ ))`

= `(1 + (sin^2θ + cos^2 θ) + 2 sin θ)/(cos θ(1 + sin θ))`

= `(1 + 1 + 2sin θ)/(cos θ(1 + sin θ))`

= `(2(1 + sin θ))/(cos θ(1 + sin θ))`

= 2 sec θ

Hence proved.